Alternating Recurrence relation $a_n = b_{n-1} + 5$ and $b_n = na_{n-1}$ I am racking my brain on solving the relation where:
$$a_n = b_{n-1} + 5$$
$$b_n = na_{n-1}$$
where $a_0$ = $b_0$ = 1
I am trying to find the closed form for $a_n$.  I have tried to shifting $b_n = na_{n-1}$ to $b_{n-1} = (n-1)a_{n-2}$ by substituting n with n-1.  Then I plug $b_{n-1}$ in to $a_n = b_{n-1} + 5$ to get $a_n = (n-1)a_{n-2} + 5$ 
I am trying to use the method of generating functions to do this.  I have that $A(x) = \sum_{n=0}^{\infty} a_{n}x^{n}$.  Now I move to the n = 3 term of the summand and get $1+6x+6x^2 + \sum_{n=3}^{\infty} a_{n}x^{n}$ as $a_0 = 1, a_1 = a_2 = 6$ Then we substitute in the relation to get $1+6x+6x^2 + x^2\sum_{n=3}^{\infty} ((n-1)a_{n-2}+5)x^{n-2}$ = $1+6x+6x^2 + x^2\sum_{n=3}^{\infty} (n-1)a_{n-2}x^{n-2}+\sum_{n=3}^{\infty}5x^{n-2}$
I know what to do with the latter term as $\sum_{n=3}^{\infty}5x^{n-2} = 5x/(1-x)$.  I however am not sure what to do with the $\sum_{n=3}^{\infty} (n-1)a_{n-2}x^{n-2}$ term as there is an (n-1) in it.  I do however see that $(n-1)a_{n-2}x^{n-2}$ looks like it could be integrated with respect to x to become $a_{n-2}x^{n-1}$.  If I then pull out a x I could end up with:
$1+6x+6x^2 + x^3/(1-x)+ 5x/(1-x)$
But I am first shaky on how I came to this, but am furthermore wondering how this would be a closed form i.e. we don't even have x's in the sequence.
Thoughts would be very appreciated.
Thanks,
Brian
 A: Start with the recursion for $a_n$ :
$$
a_n=(n-1)a_{n-2}+5
$$

Let $c_n=\dfrac{a_{2n}}{(2n-1)!!}$ and we get
$$
\begin{align}
a_{2n}&=(2n-1)a_{2n-2}+5\\
(2n-1)!!\,c_n&=(2n-1)!!\,c_{n-1}+5\\
c_n&=c_{n-1}+\frac5{(2n-1)!!}
\end{align}
$$
Thus,
$$
\begin{align}
a_{2n}
&=(2n-1)!!\left(1+5\sum_{k=1}^n\frac1{(2k-1)!!}\right)\\
&=\left\lfloor(2n-1)!!\,\left(1+5\sqrt{\frac{e\pi}{2}}\mathrm{erf}\left(\frac1{\sqrt2}\right)\right)\right\rfloor&&\text{for }n\ge3
\end{align}
$$

Let $c_n=\dfrac{a_{2n+1}}{(2n)!!}$ and we get
$$
\begin{align}
a_{2n+1}&=2na_{2n-1}+5\\
(2n)!!c_n&=(2n)!!c_{n-1}+5\\
c_n&=c_{n-1}+\frac5{(2n)!!}
\end{align}
$$
Thus,
$$
\begin{align}
a_{2n+1}
&=(2n)!!\left(6+5\sum_{k=1}^n\frac1{(2k)!!}\right)\\
&=\left\lfloor(2n)!!\,\left(1+5\sqrt{e}\right)\right\rfloor&&\text{for }n\ge2
\end{align}
$$

Thus, for indices greater than $4$, we get the closed formulae
$$
\begin{align}
a_{2n}&=\left\lfloor c_{\text{even}}\,(2n-1)!!\right\rfloor&&\text{for }n\ge3\\
a_{2n+1}&=\left\lfloor c_{\text{odd}}\,(2n)!!\right\rfloor&&\text{for }n\ge2\\
\end{align}
$$
where
$$
\begin{align}
c_{\text{even}}&=1+5\sqrt{\frac{e\pi}{2}}\mathrm{erf}\left(\frac1{\sqrt2}\right)&&=8.05343067321223998845412355710\\
c_{\text{odd}}&=1+5\sqrt{e}&&=9.24360635350064073424325393907\\
\end{align}
$$
A: We have that $a_{n+1}=na_{n-1}+5$.
Define $A(z)$ as you did $\sum a_nz^n$. Multiply by $z^{n+1}$ and add for all $n$. We get
$$\sum a_{n+1}z^{n+1}=z^2\sum na_{n-1}z^{n-1}+5\sum z^{n+1}$$
which is
$$A(z)=z^2(zA(z))'+\frac{5z}{1-z}.$$
Useful to translate operations on the sequence to operations on the generating function it this list.
You may have to add a polynomial to the equation depending on the initial conditions, which I am not looking.
Can you solve the differential equation?
There are general procedures to solve these types of differential equations.
A: Maple gets the following.  For $n$ odd.  In terms of the incomplete Gamma function.
$$
b_{{n}}=\frac{1}{4\sqrt{\pi}}{{{2}^{n/2+1/2} \left( 10\,{{\rm e}^{1/2}}\sqrt {\pi }\;
\Gamma  \left( n/2+1 \right) {{\rm erf}\left(\sqrt {2}/2\right)}
\sqrt {2}+5\,\sqrt {2}n\Gamma  \left( n/2,1/2 \right) {{\rm e}^{1/2
}}\sqrt {\pi }\\-10\,{{\rm e}^{1/2}}\sqrt {\pi }\;\Gamma  \left( n/2+1
 \right) \sqrt {2}+4\,\Gamma  \left( n/2+1 \right)  \right) }}
$$
$b_1=1, b_3=18, b_5=115$
and for $n$ even
$$
b_{{n}}= \left( 5\,{{\rm e}^{1/2}}\Gamma  \left( n/2,1/2 \right) +
\Gamma  \left( n/2 \right)  \right) {2}^{n/2-1}n
$$
$b_0=0,b_2=12,b_4=68$
A: Start with $a_n = (n − 1) a_{n − 2} + 5$, and note that this is a first order linear recurrence in $b_n = a_{2 n + 1}$:
$$
b_{n + 1} = 2 n b_n + 5
$$
Summing factor is $\prod_{0 \le k \le n} 2 n = 2^n n!$:
$$
\frac{b_{n + 1}}{2^n n!} - \frac{b_n}{2^{n - 1} (n - 1)!}
  = \frac{5}{2^n n!}
$$
Sum for $1 \le k \le n$ to get:
$$
\frac{b_{n + 1}}{2^n n!} - b_1 = 5 \sum_{1 \le k \le n} \frac{2^{-n}}{n!}
$$
The right hand side is (almost) $5 \left(e^{1/2} - 1 \right)$, so $a_{2 n + 1} = b_n \approx a_1 + 5 \cdot 2^{n - 1} (n - 1)! \left(e^{1/2} - 1\right)$
